Differential Equations

First Order Equations

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Fluid Flow from a Vessel

Solved Problems

Example 1.

Derive the differential equation of fluid leakage from a conical vessel and determine the total flow time \(T.\) The upper base radius of the conical vessel is \(R,\) the lower base radius is \(a.\) The initial height of the fluid is \(H\) (Figure \(5\)).

Solution.

Fluid flowing out of a conical vessel
Figure 5.
The cross section of the cone is a triangle
Figure 6.

The change of the fluid level at the height \(z\) is described by the differential equation

\[S\left( z \right)\frac{{dz}}{{dt}} = q\left( z \right),\]

where \(S\left( z \right)\) is the cross-sectional area at the height \(z\) and \(q\left( z \right)\) is the fluid flow depending on its height \(z.\)

Based on the geometry of the system, we can assume that the Torricelli's law holds. Therefore, one can write:

\[q\left( z \right) = - \pi {a^2}\sqrt {2gz} ,\]

where \(a\) is the radius of the opening at the bottom of the cone. Taking into account that this opening is small enough, we can consider the cross section of the cone as a triangle (Figure \(6\) above). Then it follows from the similarity of the triangles that

\[\frac{R}{H} = \frac{r}{z}.\]

Hence, the surface area of the fluid at the height \(z\) is given by

\[S\left( z \right) = \pi {r^2} = \pi {\left( {\frac{{Rz}}{H}} \right)^2} = \frac{{\pi {R^2}{z^2}}}{{{H^2}}}.\]

Substituting \(S\left( z \right)\) and \(q\left( z \right)\) into the differential equation gives:

\[\frac{{\pi {R^2}{z^2}}}{{{H^2}}}\frac{{dz}}{{dt}} = - \pi {a^2}\sqrt {2gz} .\]

After some transformations we get the following differential equation:

\[{z^{\frac{3}{2}}}dz = - \frac{{{a^2}{H^2}}}{{{R^2}}}\sqrt {2g} dt.\]

Integrate both sides taking into account that the level of the fluid decreases from its initial value \(H\) to zero for the time \(T:\)

\[\int\limits_H^0 {{z^{\frac{3}{2}}}dz} = - \int\limits_0^T {\frac{{{a^2}{H^2}}}{{{R^2}}}\sqrt {2g} dt} ,\;\; \Rightarrow \left. {\left( {\frac{{{z^{\frac{5}{2}}}}}{{\frac{5}{2}}}} \right)} \right|_0^H = \frac{{{a^2}{H^2}}}{{{R^2}}}\sqrt {2g} \left[ {\left. {\left( t \right)} \right|_0^T} \right],\;\; \Rightarrow \frac{2}{5}{H^{\frac{5}{2}}} = \frac{{{a^2}{H^2}}}{{{R^2}}}\sqrt {2g} T,\;\; \Rightarrow \frac{1}{5}\sqrt {\frac{{2H}}{g}} = \frac{{{a^2}}}{{{R^2}}}T,\;\; \Rightarrow T = \frac{{{R^2}}}{{5{a^2}}}\sqrt {\frac{{2H}}{g}} .\]

Here we can again see the analogy with the fall of a solid particle from the height \(H\) in the Earth's gravitational field. As it's known, the fall time is given by the formula:

\[T = \sqrt {\frac{{2H}}{g}}.\]

If we compare this result with the case when the fluid flows out of a cylinder, we see that under the same values of \(H, R\) and \(a\) the fluid flow time for the cone is exactly \(5\) times less than for the cylinder (though the volume of fluid in the conical vessel is only \(3\) times less!) Such integer-valued ratios in nature seem surprising, don't they?

Example 2.

Investigate the fluid flow from a thin pipe of the radius \(R\) and height \(H\) assuming that the pipe is completely filled.

Solution.

Fluid flowing out of a thin pipe
Figure 7.
Dependence of the flowing out time t on the ratio H/h
Figure 8.

Similar to the previous examples, we can write the fluid balance equation at an arbitrary height \(z\) in the form:

\[S\left( z \right)\frac{{dz}}{{dt}} = q\left( z \right).\]

In the given case the cross-sectional area \(S\left( z \right)\) is constant:

\[S\left( z \right) = S = \pi {R^2},\]

and the flow of the fluid is determined by the formula:

\[q\left( z \right) = - kz,\]

where \(k\) depends on the opening size, wettability and other parameters.

As a result we obtain the simple equation:

\[\pi {R^2}\frac{{dz}}{{dt}} = - kz,\]

or after separating variables:

\[\frac{{dz}}{z} = - \frac{k}{{\pi {R^2}}}dt.\]

Now we can integrate this equation assuming that the fluid level decreases from \(H\) to \(h\) for the time from \(0\) to \(t:\)

\[\int\limits_H^h {\frac{{dz}}{z}} = - \int\limits_0^t {\frac{k}{{\pi {R^2}}}dt} ,\;\; \Rightarrow \left. {\left( {\ln z} \right)} \right|_h^H = \frac{k}{{\pi {R^2}}}t,\;\; \Rightarrow t = \frac{{\pi {R^2}}}{k}\left( {\ln H - \ln h} \right) = \frac{{\pi {R^2}}}{k}\ln \frac{H}{h}.\]

The dependence of \(t\) on the ratio \(\frac{H}{h}\) is shown schematically in Figure \(8.\) This curve is similar to the dependence of time \(T\) on the height \(H\) for a wide cylindric vessel for which the Torricelli's law is valid. Interestingly that the flow time \(t\) formally approaches infinity as \(h \to 0\) in the given simple model.

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